calculate weight of phosphorus (P4) required which contain same number of phosphorus atoms as in a simple contain 98 gram of H3Po4, 68 gram of PH3, 568 gram of P4O10?
Answers
Answer:
3H2O3NaH2PO2 + PH3 is
Explanation:
so that weight of gram 688bl
Answer:
The weight of phosphorus (P₄) required is 341 grams.
Explanation:
Given,
A sample contains (w1 = 98) grams of H₃PO₄, (w2 = 68) grams of PH₃, and (w3 = 568) grams of P₄O₁₀.
The weight of phosphorus (w) = weight of phosphorus atoms in the given sample.
To find,
The weight of Phosphorus(P₄)
Calculation,
As the weight of phosphorus (w) = weight of phosphorus atoms in (H₃PO₄ + PH₃ + P₄O₁₀)
⇒ No. of atoms of P₄ = no. of atoms of P in (H₃PO₄ + PH₃ + P₄O₁₀)
⇒ 4 × no. of moles of P₄ × Nₐ = 1 × no. of moles of H₃PO₄ × Nₐ + 1 × no. of moles of PH₃ × Nₐ + 4 × no. of moles of P₄O₁₀ × Nₐ.
⇒ 4 × n × Nₐ = 1 × n1 × Nₐ + 1 × n2 × Nₐ + 4 × n3 × Nₐ
⇒ 4 × (w/124) = 1 × (w1/98) + 1 × (w2/54) + 4 × (w3/284)
(As G.M.W of H₃PO₄ = 98, G.M.W of PH₃ = 54, G.M.W. of P₄O₁₀ = 284)
⇒ w/31 = 98/98 + 68/34 + 4 (568/284)
⇒ w/31 = 1 + 2 + 8
⇒ w = 341 grams.
Therefore, the weight of phosphorus (P₄) required is 341 grams
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